Answer:
At t=2s the car reach its minimum speed
Minimum speed=8m/s
Explanation:
We are given that velocity of car
![v(t)=t^2-4t+12](https://img.qammunity.org/2022/formulas/mathematics/college/j05tnd42r65jx4h4001abmp32ixtn99lu6.png)
0<t<4
Differentiate w.r.t t
![v'(t)=2t-4](https://img.qammunity.org/2022/formulas/mathematics/college/466z8w64mapuvt6jzww77ffm06w5y1k3pq.png)
![v'(t)=0](https://img.qammunity.org/2022/formulas/mathematics/college/6y5ufx9lxqbecgn6jvz34w17iqjlog57oe.png)
![2t-4=0](https://img.qammunity.org/2022/formulas/mathematics/college/y2zbtbofu42o1zpq0bfwervfol7mg80vjf.png)
![2t=4](https://img.qammunity.org/2022/formulas/mathematics/college/p5lx7jhkgjqgpctdbb5gjn0lzimqbf3y1g.png)
![t=2](https://img.qammunity.org/2022/formulas/mathematics/college/5jwqdwr8ethtyb575gdq8mzxcrupgdnqbx.png)
![v''(t)=2>0](https://img.qammunity.org/2022/formulas/mathematics/college/hnrh3jgdmqm6ltold7h2ze7imt8dxmbguz.png)
The velocity is minimum at t=2 s
Substitute t=2
![v(2)=2^2-4(2)+12](https://img.qammunity.org/2022/formulas/mathematics/college/v8e03oj9bqyqecqd9os9ha4eaih2p8zf7a.png)
![v(2)=8m/s](https://img.qammunity.org/2022/formulas/mathematics/college/rrab9y5uxqarvtkn6hjjekd0j8at7q3c70.png)
Hence, the speed of car is minimum at t=2 se and minimum speed of car=8m/s